Why the Brownian Exponential Process Is a Martingale

# Steven Shreve: Stochastic Calculus and Finance


# Steven Shreve: Stochastic Calculus and Finance












The lecture notes have the following theorem:

Let $\theta\in \mathbb{R}$ be given and $B(t)$ stands for the Brownian motion which is a martingale, then $Z(t)=exp\{-\theta B(t)-\dfrac{1}{2}\theta^2t\}$ is also a martingale.

$\underline{proof:}$ Let $0\leq s\leq t$ be given. Then $$\mathbb{E}[Z(t)|\mathbb{F}(s)]=\mathbb{E}[exp\{-\theta (B(t)-B(s)) -B(s) -\dfrac{1}{2}\theta^2((t-s)+s)\}|\mathbb{F}(s)]\Rightarrow \\ \mathbb{E}[Z(t)|\mathbb{F}(s)]=Z(s)\mathbb{E}[exp\{-\theta (B(t)-B(s)) -\dfrac{1}{2}\theta^2(t-s)\}|\mathbb{F}(s)]\Rightarrow \\ \mathbb{E}[Z(t)|\mathbb{F}(s)]=Z(s)exp\{\dfrac{1}{2}(-\theta)^2\mathbb{Var}(B(t)-B(s))-\dfrac{1}{2}\theta^2(t-s)\}=Z(s)$$ where $X=B(t)-B(s)\sim N(0,t-s)$. My question is how can you proove this part $\mathbb{E}[Z(t)|\mathbb{F}(s)]=Z(s)exp\{\dfrac{1}{2}(-\theta)^2\mathbb{Var}(B(t)-B(s))-\dfrac{1}{2}\theta^2(t-s)\}$ analytically by using the normal pdf. I am missing something in my effort to proove this part, because no textbook from those that I have does it analytically.

## Answer by Kevin (score 6, accepted)

https://quant.stackexchange.com/a/50788

Note merely that $B_t=B_s+(B_t-B_s)$ which is the sum of independent normally distributed random variables. In particular, $B_s$ is $\mathbb{F}_s$-measurable and $B_{t-s}$ is independent of $\mathbb{F}_s$. Thus, \begin{align*} \mathbb{E}_s[Z_t] &= \mathbb{E}_s\left[\exp\left(-\frac{1}{2}\theta^2t+\theta B_t\right)\right] \\ &= \mathbb{E}_s\left[\exp\left(-\frac{1}{2}\theta^2t+\theta B_s+\theta B_{t-s}\right)\right] \\ &= \exp\left(-\frac{1}{2}\theta^2s\right)\exp\left(-\frac{1}{2}\theta^2(t-s)\right)\cdot\mathbb{E}_s\left[\exp\left(\theta B_s+\theta B_{t-s}\right)\right]\\ &= \exp\left(-\frac{1}{2}\theta^2s\right)\cdot\exp\left(-\frac{1}{2}\theta^2(t-s)\right)\cdot\exp\left(\theta B_s\right)\cdot\mathbb{E}\left[\exp\left(\theta B_{t-s}\right)\right]\\ &= Z_s\exp\left(-\frac{1}{2}\theta^2(t-s)\right)\cdot\exp\left(\frac{1}{2}\theta^2 \mathrm{Var}[B_{t-s}]\right) \\ &= Z_s\exp\left(-\frac{1}{2}\theta^2(t-s)+\frac{1}{2}\theta^2 (t-s)\right) \\ &= Z_s. \end{align*}

And you're done and have shown that $(Z_t)$ is a martingale (with respect to the natural filtration of the Brownian motion).

You don't need to integrate the normal density. That's just tedious. You should first decompose the Brownian motion as mentioned in the beginning. The $-\frac{1}{2}\theta^2t$ can be split up similarly and pulled out of the expectation as it is non-random. Then, you can use measurability and independence (two key properties of the conditional expectation) to pull out $B_s$ and reduce the conditional expectation of $B_{t-s}$ to a simple expectation. Then, note that $B_{t-s}\sim N(0,t-s)$ and that $\mathbb{E}\left[e^{m+sZ}\right]=e^{m+0.5s^2}$ - a useful formula to keep in mind.